# lim_(n->oo)prod_(k=2)^n(1-1/k^2)?

Jul 13, 2018

${\prod}_{k = 2}^{\infty} \left(1 - \frac{1}{k} ^ 2\right) = \frac{1}{2}$

#### Explanation:

Consider the sequence of integrals:

${I}_{m} = {\int}_{0}^{\frac{\pi}{2}} {\sin}^{m} x \mathrm{dx}$

As the integrand is defined and continuous in the whole interval, these integrals exist for every $m \in \mathbb{N}$.

Evaluate now the integral by parts:

${I}_{m} = - {\int}_{0}^{\frac{\pi}{2}} {\sin}^{m - 1} x d \left(\cos x\right)$

${I}_{m} = - {\left[{\sin}^{m - 1} x \cos x\right]}_{0}^{\frac{\pi}{2}} + {\int}_{0}^{\frac{\pi}{2}} \left(m - 1\right) {\sin}^{m - 2} x {\cos}^{2} x \mathrm{dx}$

${I}_{m} = \left(m - 1\right) {\int}_{0}^{\frac{\pi}{2}} {\sin}^{m - 2} x \left(1 - {\sin}^{2} x\right) \mathrm{dx}$

${I}_{m} = \left(m - 1\right) {I}_{m - 2} - \left(m - 1\right) {I}_{m}$

and finally we have the recursive formula:

${I}_{m} = \frac{m - 1}{m} {I}_{m - 2}$

so that:

${I}_{m - 2} / {I}_{m} = \frac{m}{m - 1}$

and:

${\lim}_{m \to \infty} {I}_{m - 2} / {I}_{m} = 1$

Given that:

${I}_{0} = {\int}_{0}^{\frac{\pi}{2}} \mathrm{dx} = \frac{\pi}{2}$

${I}_{2} = {\int}_{0}^{\frac{\pi}{2}} {\sin}^{2} x \mathrm{dx} = \frac{\pi}{4}$

consider now for every $m$ the ratio:

${I}_{m - 1} / {I}_{m + 1} = {I}_{m - 1} / {I}_{m} {I}_{m} / {I}_{m + 1}$

and evaluate each ratio using the recursive formula:

I_(m-1)/I_(m) = ((m-2)/(m-1) I_(m-3))/((m-1)/(m) I_(m-2)

${I}_{m - 1} / {I}_{m} = \left(\frac{m \left(m - 2\right)}{m - 1} ^ 2\right) {I}_{m - 3} / {I}_{m - 2}$

${I}_{m - 1} / {I}_{m} = \left(\frac{\left(m - 1 + 1\right) \left(m - 1 - 1\right)}{m - 1} ^ 2\right) {I}_{m - 3} / {I}_{m - 2}$

${I}_{m - 1} / {I}_{m} = \left(\frac{{\left(m - 1\right)}^{2} - 1}{m - 1} ^ 2\right) {I}_{m - 3} / {I}_{m - 2}$

${I}_{m - 1} / {I}_{m} = \left(1 - \frac{1}{m - 1} ^ 2\right) {I}_{m - 3} / {I}_{m - 2}$

and:

I_(m)/I_(m+1)= ((m-1)/(m) I_(m-2))/((m)/(m+1) I_(m-1)

${I}_{m} / {I}_{m + 1} = \left(\frac{\left(m + 1\right) \left(m - 1\right)}{m} ^ 2\right) {I}_{m - 2} / {I}_{m - 1}$

${I}_{m} / {I}_{m + 1} = \left(\frac{{m}^{2} - 1}{m} ^ 2\right) {I}_{m - 2} / {I}_{m - 1}$

${I}_{m} / {I}_{m + 1} = \left(1 - \frac{1}{m} ^ 2\right) {I}_{m - 2} / {I}_{m - 1}$

Then:

${I}_{m - 1} / {I}_{m + 1} = \left(1 - \frac{1}{m} ^ 2\right) {I}_{m - 2} / {I}_{m - 1} \left(1 - \frac{1}{m - 1} ^ 2\right) {I}_{m - 3} / {I}_{m - 2}$

${I}_{m - 1} / {I}_{m + 1} = \left(1 - \frac{1}{m} ^ 2\right) \left(1 - \frac{1}{m - 1} ^ 2\right) {I}_{m - 3} / {I}_{m - 1}$

and iterating:

${I}_{m - 1} / {I}_{m + 1} = \left(1 - \frac{1}{m} ^ 2\right) \left(1 - \frac{1}{m - 1} ^ 2\right) \left(1 - \frac{1}{m - 2} ^ 2\right) \left(1 - \frac{1}{m - 3} ^ 2\right) {I}_{m - 5} / {I}_{m - 3}$

or:

${I}_{m - 1} / {I}_{m + 1} = {\prod}_{j = 0}^{p} \left(1 - \frac{1}{m - j} ^ 2\right) {I}_{m - p - 2} / {I}_{m - p}$

So for $p = m - 2$ we have:

${I}_{m - 1} / {I}_{m + 1} = {\prod}_{j = 0}^{m - 2} \left(1 - \frac{1}{m - j} ^ 2\right) {I}_{0} / {I}_{2}$

${I}_{m - 1} / {I}_{m + 1} = {\prod}_{j = 0}^{m - 2} \left(1 - \frac{1}{m - j} ^ 2\right) \frac{\frac{\pi}{2}}{\frac{\pi}{4}}$

${I}_{m - 1} / {I}_{m + 1} = 2 {\prod}_{j = 0}^{m - 2} \left(1 - \frac{1}{m - j} ^ 2\right)$

Change now the index to $k = m - j$:

${I}_{m - 1} / {I}_{m + 1} = 2 {\prod}_{k = 2}^{m} \left(1 - \frac{1}{k} ^ 2\right)$

so:

${\prod}_{k = 2}^{m} \left(1 - \frac{1}{k} ^ 2\right) = \frac{1}{2} {I}_{m - 1} / {I}_{m + 1}$

and passing to the limit:

${\prod}_{k = 2}^{\infty} \left(1 - \frac{1}{k} ^ 2\right) = \frac{1}{2} {\lim}_{m \to \infty} {I}_{m - 1} / {I}_{m + 1} = \frac{1}{2}$

Jul 14, 2018

${\lim}_{n \to \infty} {\prod}_{k = 2}^{n} \left(1 - \frac{1}{k} ^ 2\right)$

$= {\lim}_{n \to \infty} {\prod}_{k = 2}^{n} \frac{\left(k - 1\right) \left(k + 1\right)}{k} ^ 2$

$= {\lim}_{n \to \infty} {\underbrace{\frac{\left(1\right) \left(3\right)}{2} ^ 2}}_{k = 2} \cdot {\underbrace{\frac{\left(2\right) \left(4\right)}{3} ^ 2}}_{k = 3} \cdot {\underbrace{\frac{\left(3\right) \left(5\right)}{4} ^ 2}}_{k = 4} \cdot {\underbrace{\frac{\left(4\right) \left(6\right)}{5} ^ 2}}_{k = 5} \ldots \cdot {\underbrace{\frac{\left(n - 3\right) \left(n - 1\right)}{n - 2} ^ 2}}_{k = n - 2} \cdot {\underbrace{\frac{\left(n - 2\right) \left(n\right)}{n - 1} ^ 2}}_{k = n - 1} \cdot {\underbrace{\frac{\left(n - 1\right) \left(n + 1\right)}{n} ^ 2}}_{k = n}$

$= {\lim}_{n \to \infty} {\underbrace{\frac{\left(1\right)}{2}}}_{k = 2} \cdot {\underbrace{\frac{\left(1\right) \left(1\right)}{1} ^ 2}}_{k = 3} \cdot {\underbrace{\frac{\left(1\right) \left(1\right)}{1} ^ 2}}_{k = 4} \cdot \underbrace{{\underbrace{\frac{\left(1\right) \left(6\right)}{5}}}_{k = 5} \cdot \ldots \cdot {\underbrace{\frac{\left(n - 3\right) \left(1\right)}{n - 2}}}_{k = n - 2}} {\setminus}_{\text{ Not fully factored }} \cdot {\underbrace{\frac{\left(1\right) \left(1\right)}{1} ^ 2}}_{k = n - 1} \cdot {\underbrace{\frac{\left(1\right) \left(n + 1\right)}{n}}}_{k = n}$

The pattern : every term, in conjunction with the immediately preceding and succeeding terms, reduces to $1$. This is not true for the first and last term which are as shown.

$= {\lim}_{n \to \infty} \frac{1}{2} \cdot 1 \cdot 1 \cdot \ldots \cdot 1 \cdot \frac{n + 1}{n}$

$= \frac{1}{2} {\lim}_{n \to \infty} 1 + \frac{1}{n} = \frac{1}{2}$